yy’-e^x=0, and y=4 when x=0, which means that:

Question

y=x-lnx^2+4
y^2=4x^2+3
y=-2x+1/2x^3
y^2=2e^x+14
1/2ln|x^2+4|+6

Answer 1

Hmmm. Not so fast. I have no idea where that solution came from.
y y' = e^x
1/2 y^2 = e^x + c
y^2 = 2e^x + c
since y(0) = 4, 2+c = 16 ==> c = 14 and so
y^2 = 2e^x + 14
y = √2 (e^x + 7)